1. ## Solving Trig functions?

I'm not sure how to solve these trig functions;

1. 2sin(theta)−sqrt(2)tan(theta)= 0 ; 0 ≤ theta < 2pi

2. sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 ; 0 ≤ theta < 2pi
for #2 i factored out the sin(theta) (cos(theta)-1)- cos(theta)+1 =0
then moved the - cos(theta)+1 to the right side and solved for sin(theta)
is that right so far? D:

2. 1. Multiply through by $\displaystyle ]\cos(\theta)$ and solve by factoring

3. Originally Posted by dondonlouie
2. sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 ; 0 ≤ theta < 2pi
for #2 i factored out the sin(theta) (cos(theta)-1)- cos(theta)+1 =0
then moved the - cos(theta)+1 to the right side and solved for sin(theta)
is that right so far? D:
It's probably better to rewrite that first line as
$\displaystyle sin(\theta) \cdot (cos(\theta) - 1) - (cos(\theta) - 1) = 0$

Can you see where to go from here?

-Dan

4. Originally Posted by topsquark
It's probably better to rewrite that first line as
$\displaystyle sin(\theta) \cdot (cos(\theta) - 1) - (cos(\theta) - 1) = 0$

Can you see where to go from here?

-Dan
wouldn't the cos(theta)-1 cancel out leaving sin(theta)=0?

5. Originally Posted by dondonlouie
wouldn't the cos(theta)-1 cancel out leaving sin(theta)=0?
Careful! When you have the equation ab = 0 then either a = 0 or b = 0. So what this equation is saying:
$\displaystyle (sin(\theta) - 1)(cos(\theta) - 1) = 0 \implies sin(\theta) - 1 = 0 \text{ or } cos(\theta) - 1 = 0$

-Dan

6. Originally Posted by topsquark
Careful! When you have the equation ab = 0 then either a = 0 or b = 0. So what this equation is saying:
$\displaystyle (sin(\theta) - 1)(cos(\theta) - 1) = 0 \implies sin(\theta) - 1 = 0 \text{ or } cos(\theta) - 1 = 0$

-Dan
how did you get from sin(theta)(cos(theta)-1)-(cos(theta)-1)= 0
to (sin(theta)-1)(cos(theta)-1)?

7. Take out $\displaystyle \displaystyle (\cos{\theta} - 1)$ as a factor...

8. Originally Posted by Prove It
Take out $\displaystyle \displaystyle (\cos{\theta} - 1)$ as a factor...
ohhh okay i see it
thank you so much for helping me

9. Originally Posted by dondonlouie
ohhh okay i see it